Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $y = \dfrac{4(2z + 5)}{7z} \div \dfrac{16z + 40}{5z} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{4(2z + 5)}{7z} \times \dfrac{5z}{16z + 40} $ When multiplying fractions, we multiply the numerators and the denominators. $y = \dfrac{ 4(2z + 5) \times 5z } { 7z \times (16z + 40) } $ $ y = \dfrac {5z \times 4(2z + 5)} {7z \times 8(2z + 5)} $ $ y = \dfrac{20z(2z + 5)}{56z(2z + 5)} $ We can cancel the $2z + 5$ so long as $2z + 5 \neq 0$ Therefore $z \neq -\dfrac{5}{2}$ $y = \dfrac{20z \cancel{(2z + 5})}{56z \cancel{(2z + 5)}} = \dfrac{20z}{56z} = \dfrac{5}{14} $